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2

Why are you restricting yourself to $x \in (0,\infty)$ when the domain is for the entire real line? In this case, it seems easier to go with Fourier transforms, i.e., $$u(s,t) = \int_{-\infty}^{\infty} dx \, U(x,t) \, e^{i s x}$$ Then the PDE becomes $$u_t + s^2 u = \sqrt{2 \pi} \, e^{-s^2/2} = e^{-s^2 t}\frac{d}{dt}\left [e^{s^2 t} u \right ]$$ ...


2

Let me answer a different more general question $$ v(t,S) = \mathrm{e}^{-rt}U\left(t,g(x,t)\right).\tag{1} $$ $$ v_t = -r\mathrm{e}^{-rt}U\left(t,g(x,t)\right) + \mathrm{e}^{-rt}U_t.\tag{2} $$ now you must be happy with Eq.(2)? so lets compute $U_t$, $$ U_t = \dfrac{\partial U}{\partial t} + \dfrac{\partial U}{\partial g}\dfrac{\partial g(x,t)}{\partial ...


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The change of variables should make one of the new variables constant on the characteristic lines. Thus try $w = x - y$, $z = y$.


2

Of course it is! The method of characteristics for your linear 1st oder PDE reads: $$\frac{\mathrm{d}x}{1} = \frac{\mathrm{d}y}{1}= \frac{\mathrm{d}u}{u},$$ which leads to $x-y = \xi$ (from 1st and 2nd identities) and $u = \eta \, e^x$ (from 1st and 3rd identities). Put $\eta = \eta(\xi ) = \eta(x-y)$ and you are done. Note the following funny fact of ...


1

$ \newcommand{\pd}[2]{ \frac{\partial #1}{\partial #2} } $ Attending to the positive answer to my request from the OP, consider the non-linear 1st order PDE: $$F(x_i,u,p_i) = 0, \quad i = 1,\ldots, n, \quad p_i = \pd{u}{x_i}, \quad u = u(x_1,\ldots,x_n). \tag{1}$$ Assume $F \in \mathcal{C}^1_{x_i}$ and compute the partial derivative of eq. $(1)$ with ...


1

No, the conjugate of a harmonic function that is continuous up to the boundary need not be continuous up to the boundary, or even bounded. This is related to the fact that the Hilbert transform does not preserve continuity (though it does preserve Hölder continuity of exponents $\alpha\in (0,1)$). Here is an example. Let $F$ be a conformal map of the ...


1

In order to derive the Black-Scholes equation, you can do it in a more intuitive way (with economics). Also, this needs understanding of Brownian Motions and stochastic differential equations (SDEs), Îto's Lemma, and some basical assumptions that I will not list here (they're already available in Wikipedia and other references). Let's see it in two steps: ...


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There's a simpler explanation of why $R'(0) = 0$: You assumed your solution is radially symmetric, and performed a separation of variables. If $R'(0) \neq 0$ you have a cusp at the origin, where the function $u$ is in fact not differentiable. So you require $R'(0) = 0$. The power series explanation is basically the same: you need $$ R'' + R'/r = c R $$ ...


1

You remembered to use the chain rule, but apparently forgot the product rule. Use it once: $$ V_x=e^{-ax-by}U_x-a e^{-ax-by}U $$ then use it again: $$ V_{xx}=e^{-ax-by}U_{xx}- a e^{-ax-by}U_x - a e^{-ax-by}U_x + a^2 e^{-ax-by}U $$ You may notice the same term appearing in the middle twice; it saves a bit of time to remember the product formula for second ...


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In fact, $\alpha\beta$ is optimal. To prove it, let $f,g:[0,1]\to\mathbb{R}$ with $f(x)=x^\alpha$ and $g(x)=x^\beta$. Note that $f\circ g\in C^{\alpha\beta}$, however, for all $\gamma>\alpha\beta$, $f\circ g$ does not belong to $C^\gamma$.


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Using abstract index notation: There are two types of indices here: I will use $i,j,\ldots$ for indices relative to the domain (which appears to be $\mathbb{R}^2$) and $A,B,\ldots$ for indices relative to the co-domain (basically from the number of equations) which is $\mathbb{R}^N$. We can write $\mathbf{c}$ as the rank 4 tensor in index notation $$ ...



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